# Thread: help with algebra in calc question: infinite limit

1. ## help with algebra in calc question: infinite limit

i was trying to find the limits of a function when i came onto a problem

square root of ( (x^2) + 9x + 3)
multiply by (1/x)

this should simplify to:

square root of ((x) + (9/x) + (3 / (x^2) )

how does this work?

i was trying to slove this problem

find the limit of

((square root of ((x^2) + 9x + 3)) minus x

lim when x approaches infinity

just convert the 1/x into square root of (1/ (x^2)
this makes it possible to simplify the stuff under the main square root.
this is why x^2 simplifies to 1, 9x simplifies to 9/x, etc.

thanks to me?

2. Originally Posted by jedward
just convert the 1/x into square root of (1/ (x^2)
this makes it possible to simplify the stuff under the main square root.
this is why x^2 simplifies to 1, 9x simplifies to 9/x, etc.

thanks to me?
as it stands, you cannot multiply by 1/x, unless there is some part of the question you are not writing, in which case you can multiply by something like $\frac {\frac 1x}{\frac 1x}$, if it were a rational function, but it depends.

$\lim_{x \to \infty} \sqrt{x^2 + 9x + 3} = \infty$ that was your first question

the second is different. you would multiply by the conjugate over itself to find the limit

$\lim_{x \to \infty} \sqrt{x^2 + 9x + 3} - x = \lim_{x \to \infty} ( \sqrt{x^2 + 9x + 3} - x) \cdot \frac {\sqrt{x^2 + 9x + 3} + x}{\sqrt{x^2 + 9x + 3} + x}$

now simplify and finish up. you will need to note the following: $\sqrt{x^2 + 9x + 3} = \sqrt{x^2 \left(1 + \frac 9x + \frac 3{x^2} \right)} = \sqrt{x^2} \sqrt{1 + \frac 9x + \frac 3{x^2}} = |x| \sqrt{1 + \frac 9x + \frac 3{x^2}}$

3. Hello, jedward!

$\lim_{x\to\infty}\bigg[\sqrt{x^2+9x+3} - x\bigg]$
Multiply top and bottom by the conjugate:

. . $\frac{\sqrt{x^2+9x+3} - x}{1}\;\cdot\;\frac{\sqrt{x^2+9x+3} + x}{\sqrt{x^2+9x+3} + x} \;=\;\frac{(x^2+9x+3) - x^2}{\sqrt{x^2+9x+3} + x} \;=\;\frac{9x+3}{\sqrt{x^2+9x+3} + x}$

Divide top and bottom by $x$:

. . $\frac{\dfrac{9x}{x} + \dfrac{3}{x}} { \dfrac{\sqrt{x^2+9x+3}}{x} + \dfrac{x}{x}} \;=\;\frac{9+\dfrac{3}{x}} {\dfrac{\sqrt{x^2+9x+3}}{\sqrt{x^2}} + 1}$ $= \;\frac{9+\dfrac{3}{x}}{\sqrt{\dfrac{x^2+9x+3}{x^2 }} + 1}$

. . . . $=\;\frac{9+\dfrac{3}{x}}{\sqrt{\dfrac{x^2}{x^2} + \dfrac{9x}{x^2} + \dfrac{3}{x^2}} + 1} \;=\;\frac{9+\dfrac{3}{x}}{\sqrt{1 + \dfrac{9}{x} + \dfrac{3}{x^2}} + 1}$

Therefore: . $\lim_{x\to\infty}\left[\frac{9+\dfrac{3}{x}}{\sqrt{1 + \dfrac{9}{x} + \dfrac{3}{x^2}} + 1}\right] \;=\; \frac{9 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{9}{2}$

Edit: Jhevon beat me to it . . .
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